Timeline for Is there any sequence $a_n$ of nonnegative numbers for which $\sum_{n \geq 1}a_n^2 <\infty$ and $\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty$?

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Nov 7, 2017 at 11:09	comment	added	C.S.		I don't think we need Tensor product of operators. Call the expression $f(x_{n})$. It suffices to find $\{x_{n}\}$'s with $\|x_{n}\|=1$ and $f(x_n)$ unbounded. For a large $N$, let $x_{N}= N^{-\frac{N}{2}}$ if $N=\prod_{i=1}^{N} p_{i}^{a_{i}}$ where $a_{i}<N$ (where $p_i$ is the i-th prime) and $0$ otherwise.
Jun 18, 2011 at 20:46	vote	accept	a_MSE_user
Jun 11, 2011 at 20:08	comment	added	Noam D. Elkies		@Eric - thanks! The tensor-product description turns out not to be explicitly used in the proof (which indeed generalizes to some operators, such as the multiplicative convolution with $1/(k \log k)$, that do not factor as tensor products), but it did suggest where to find counterexamples.
Jun 11, 2011 at 20:06	history	edited	Noam D. Elkies	CC BY-SA 3.0	Changed $k$ to $m$ in the last paragraph to remove notational conflict
Jun 11, 2011 at 18:30	comment	added	Eric Naslund		+1, very nice. I really like the way you look at it as a tensor product of operators.
Jun 11, 2011 at 13:31	history	edited	Noam D. Elkies	CC BY-SA 3.0	Corrected typo noted by A.Amit; spelled out argument for $T\alpha=M\alpha$
Jun 11, 2011 at 13:23	comment	added	Noam D. Elkies		@Alon Amit: yes, thanks. I'll make the edit.
Jun 11, 2011 at 6:50	comment	added	Alon Amit		In line 3, did you mean "the sequence whose $n$-th term is..."?
Jun 11, 2011 at 4:41	history	answered	Noam D. Elkies	CC BY-SA 3.0